Homework Help:
Finding the current through an icosahedron of resistors

Consider an icosahedron all of whose edges are 2 ohm resistors. If a 12 V battery is connected in opposite vertices, what is the current flow through each resistor?

2. Relevant equations

v=ir

3. The attempt at a solution

I found somewhere that the equivalent resistance of the icosahedron would be R/2. So i use v=ir to find i=v/r =12/1=12amps and since there are 30 edges in the shape, i take 12/30 because of symmetry so the current through each resistor is 3/5amps?

Im not sure how to actually reduce a 3D shape of resistors, it has been a while since ive taken a class on circuits. Any help would be much appreciated.

I found somewhere that the equivalent resistance of the icosahedron would be R/2. So i use v=ir to find i=v/r =12/1=12amps and since there are 30 edges in the shape, i take 12/30 because of symmetry so the current through each resistor is 3/5amps?

Im not sure how to actually reduce a 3D shape of resistors, it has been a while since ive taken a class on circuits. Any help would be much appreciated.

Welcome to PF.

Not quite. Taking 12A/30 would work if all 30 resistors were simply wired in parallel, but that is not the case here. Moreover, it's not even certain that all 30 resistors have the same current.

To start, think about one of the vertexes that is connected to the battery, and try to answer these questions:
1. How many resistors are connected to that vertex?
2. How does the 12A current get divided up among those resistors?

So each of the vertices has 5 resistors connected to it. Which means if a voltage source was connected to one of these vertices, then 5/12 amps would flow through each resistor. Then each of those resistors is connected to 4 more. but of those 4, two are connected to each other so they cant have current flowing through them. Does that mean that of the 30 total resistors because of symmetry 10 of the resistors don't have any current flowing through them?
If that was true then the remaining 10 would have to have (1/2) of (5/12)?

so 10 of the resistors have 5/12 amps going through them; 10 have 0 and the other 10 have 5/24 amps?

are the 5 resistors touching any 1 vertex in parallel? so the opposite vertex would have the same situation. which would leave 10 resistors in the middle also connected in parallel? then these 3 groups would all be connected in series? so if what i am thinking is true then the total resistance would be

2/5 +2/10 +2/5=1ohm?

but this brings my back to the first conclusion that i had, so i must be doing something wrong, because if Reff=1. then the total current would be 12amps, which is what i had before. unless the current through each resistor is just 12/5. which makes sense for the entrance 5, and exit 5. are the rest of them all the same?